Ncert Solutions Chemistry Class 11th

Laboratory preparation of carbon monoxide:

Formic acid is dehydrated with concentrated sulphuric acid at 373 K.

HCOOH → H₂O + CO↑

Commercial preparation of CO:

Steam is passed over hot coke.

C + H₂O → CO + H₂

Laboratory preparation of carbon dioxide:

Calcium carbonate reacts with dilute HCl to form carbon dioxide.

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

Industrial preparation of carbon dioxide:

Limestone is heated to produce carbon dioxide.

CaCO₃ →→CaO + CO₂

The balanced equations are given below:

(i) 2BF₃ + 6LiH → B₂H₆ + 6LiF

(ii) B₂H₆ + 6H₂O → 2H₃BO₃ (orthoboric acid)

(iii) 2NaH + B₂H₆ →2Na [BH₄] (sodium borohydride)

(iv) H₃BO₃ →HBO₂ (metaboric acid) + H₂O
     4HBO₂ →H2B4O7→ 2B2O3 (boron trioxide) + H₂O

(v) Al+3NaOH→Al (OH)₃ + 3Na
 3 B₂H₆ + 6NH₃ → 2B₃N₃H₆ + 12H₂

The compounds X, Y and Z are borax, sodium metaborate + boric anhydride and boric acid respectively.

When borax is heated, it first swells and then forms a transparent glass like bead of sodium meta borate and boric anhydride.

Na₂B₄O₇ à2NaBO₂+B₂O₃+10H₂O

(Borax)       (sodium metaborate) (Boric anhydride)

Aqueous solution of borax is alkaline due to formation of strong base NaOH.

Hence, it turns red litmus blue.

Na₂B₄O₇ +7 H₂O → 4H₃BO₃ + 2NaOH

Borax reacts with sulphuric acid to form boric acid and sodium sulphate.

Na₂B₄O₇ + H₂SO₄ + 5 H₂O → 4H₃BO₃ + Na₂SO₄

(a) Inert pair effect: When the pair of electrons in the valence shell does not take part in bond formation, then this effect is called as inert pair effect.

(b)Allotropy: It is the property of the element by which an element can exist in two or more forms which have same chemical properties but different physical properties due to their structures.

(c)Catenation: The tendency to link with one another through covalent bonds to form chains and rings. This property is called catenation.
For example, carbon forms chains with (C-C) single bonds and also with multiple bonds (C = C or C = C).

The compounds X, A, B, C and D are aluminium, aluminium hydroxide, sodium tetrahydrozoaluminate (III), aluminium chloride and alumina.

Aluminium reacts with NaOH to form white PPT of Al (OH)₃→.

2Al + 3NaOH → Al (OH)₃→ + 3Na⁺

Al (OH)₃→ reacts with NaOH to form Na⁺ [Al (OH)₄→]^→.

Al (OH)₃ + NaOH → Na⁺ (Al (OH)₄)^→
Al (OH)₃→ reacts with HCL to form AlCl₃→.

Al (OH)₃ + 3HCl → AlCl₃ + 3H₂O

When Al_2→O₃→ is heated,  Al_2→O₃→ is obtained. 

2Al (OH)₃ → Al₂O₃ + 3H₂O

Tl belongs to group 13 and shows both the oxidation state +1 and +3 due to inert pair effect. Tl forms basic oxide like group I elements. TlO₂ is strongly basic.

1.14. S.I. unit of mass is kilogram (kg).

It is defined as the mass of platinum-iridium block stored at international bureau of weights and measures in France.

(a) Neutral — CO
Acidic — B₂O₃, SiO₂, CO₂ 

Basic — Tl₂O₃ 

Amphoteric — Al₂O₃, PbO₂

(b)-CO does not react with acid as well as base at room temperature.
Being acidic B₂O₃, SiO₂ and CO₂ react with alkalis to form salts.

B₂O₃ + 2NaOH à 2NaBO₂ + H₂O

SiO₂ + NaOH à 2Na₂SiO₃ + H₂O

CO₂ + 2NaOH à Na₂CO₃ + H₂O

Being Amphoteric, Al₂O₃, PbO₂ react with acids and bases.

Al₂O₃ + 2NaOH à 2NaAlO₂ + H₂O

Al₂O₃ + 3H₂SO₄à (Al₂SO₄)₃+ 3H₂O

PbO₂ + 2 NaOH à Na₂PbO₃ + H₂O

2PbO₂ + 2 H₂SO₄à 2PbSO₄ + 2H₂O + O₂

Being basic Tl₂O₃ dissolves in acids

Tl₂O₃ + 6HCl à 2TlCl₃ + 3H₂O

1.13. Pressure= Force/Area

But weight = m x g, where m = mass (in kg) and g = 9.8 m/s²

Therefore, Pressure = 1034 g cm^-2 x 9.8 m/s²

= 1034 x 10^-3 kg x (100 m)² x 9.8 m/s²

= 101332 Pa

= 1.01332 x 10⁵ Pa

1.12. Density of methanol = 0.793 kg/L, molar mass of methanol (CH₃OH) = 32g/mol = 0.032 kg/mol

V₁ =? , V₂ = 2.5 L, M₂ = 0.25 M

We can apply the formula of

M₁V₁ = M₂V₂

Or V₁ = M₂V₂/M₁

Substituting M₁ = density / molar mass, we get

M₁ = 0.793/0.032 = 24.78

V₁ = 0.25 x 2.5 / 24.78 = 0.02522 L = 25.22 mL